The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(n^1, INF).

(0) Obligation:

Runtime Complexity TRS:
The TRS R consists of the following rules:

sum(0) → 0
sum(s(x)) → +(sqr(s(x)), sum(x))
sqr(x) → *(x, x)
sum(s(x)) → +(*(s(x), s(x)), sum(x))

Rewrite Strategy: FULL

(1) CpxTrsToCpxRelTrsProof (BOTH BOUNDS(ID, ID) transformation)

Transformed TRS to relative TRS where S is empty.

(2) Obligation:

Runtime Complexity Relative TRS:
The TRS R consists of the following rules:

sum(0) → 0
sum(s(x)) → +(sqr(s(x)), sum(x))
sqr(x) → *(x, x)
sum(s(x)) → +(*(s(x), s(x)), sum(x))

S is empty.
Rewrite Strategy: FULL

(3) SlicingProof (LOWER BOUND(ID) transformation)

Sliced the following arguments:
sqr/0
*/0
*/1

(4) Obligation:

Runtime Complexity Relative TRS:
The TRS R consists of the following rules:

sum(0) → 0
sum(s(x)) → +(sqr, sum(x))
sqr → *
sum(s(x)) → +(*, sum(x))

S is empty.
Rewrite Strategy: FULL

(5) DecreasingLoopProof (EQUIVALENT transformation)

The following loop(s) give(s) rise to the lower bound Ω(n¹):
The rewrite sequence
sum(s(x)) →⁺ +(sqr, sum(x))
gives rise to a decreasing loop by considering the right hand sides subterm at position [1].
The pumping substitution is [x / s(x)].
The result substitution is [ ].

(6) BOUNDS(n^1, INF)